I'm supposed to calculate the capacitance of an unknown capacitor in series, but I'm not sure exactly which equation to use.

I know the voltage across the resistor (Vr), voltage across the capacitor (Vc), the total voltage (V), the resistance value of the resistor (R), the frequency (f), and since I was using an oscilloscope, the change in time between Vr and V. I know there's an equation that deals with these values (it includes natural number e), but I can't find it, and even if I had it I'm not exactly sure how I would rederive it to solve for the capacitance. How would I attempt this problem?

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    $\begingroup$ You are trying to calculate the capacitance of an unknown resistor? I think you mistyped? $\endgroup$ – DJBunk Mar 29 '13 at 23:43
  • $\begingroup$ @Mia You need to tell us the given information in the question, and what you need to calculate. $\endgroup$ – JKL Mar 29 '13 at 23:50
  • $\begingroup$ Shoot I did... sorry I'm really tired $\endgroup$ – Genevieve Mar 29 '13 at 23:58
  • $\begingroup$ Also you don't need the actual values right? I can't find the equation that I need. $\endgroup$ – Genevieve Mar 29 '13 at 23:59


It is not clear whether $V_c$, $V_R$ and $V$ are AC voltages or transient values in a DC circuit. I will assume that the capacitor–resistor system are supplied by an AC voltage $V_{max}$ of frequency $f$. Then the equation holds

$V=I(R-\frac{j}{\omega C})$ where $j=\sqrt{-1}$.

So that treating the circuit as a potential divider one can write

$V_c=\frac{V}{R-\frac{j}{\omega C}}\frac{1}{j\omega C}$


$V_c=\frac{V}{j\omega RC+1}$

or in terms of the magnitudes

$V_{cmax}=\frac{V_{max}}{\sqrt{1+(\omega RC)^2}}$

Therefore, knowing the maximum values for $V_c$ and $V$ you can solve this equation for C

$C=\frac{1}{\omega R}\sqrt{(V/V_c)^2-1}$ where $\omega=2\pi f$

I hope this helps.

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  • $\begingroup$ It wasn't the exact equation I needed, but it helped me find the right equation I did need, so thanks! :D $\endgroup$ – Genevieve Mar 30 '13 at 19:00

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